A curve 'C' with negative slope through the point(0,1) lies in the I Quadrant. The tangent at any point 'P' on it meets the x-axis at 'Q'. Such that `PQ=1`. Then
The curve in parametric form is

Options

A. `x=costheta+log_(e)tan(theta//2), y=sintheta`
B. `x=-costheta+log_(e)tan(theta//2),y=sintheta`
C. `x=-costheta-log_(e)tantheta//2,y=sintheta`
D. None of these

Answer

C

Explanation

Tangent at any point `P(x,y)` on curve 'C' is
`Y-y=(dy)/(dx)(X-x)`
`therefore Q=(x-(y)/((dy)/(dx)),0)`
Given, `PQ^(2)=1`
`rArr (y/y^('))^(2)+y^(2)=1`
`rArr (dy)/(dx) = -y/sqrt(1-y^(2))`…………..(1)
Putting `y=sintheta`, we get
`rArr (d(sintheta))/(dx) = (-sintheta)/(costheta)`
`rArr costheta((d)theta)/(dx) = -(sintheta)/(costheta)`
`rArr (cos^(2)theta)/(sintheta) (d)theta=-dx`
`rArr (1-sin^(2)theta)/(sintheta) =-dx`
`rArr int("cosec "theta-sintheta)(d)theta=-intdx`
`rArr x+c=-costheta-log tan(theta/2)`
Given, `y=1`, where `x=0`
`rArr x=0, theta=pi/2 rArr c=0`
The curve in parametric form is
`x=-costheta-log_(e) tan(theta)/(2), y=sintheta`